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Câu hỏi: Cho các số thực $a$, $b$ thỏa mãn $a>b>1$ và $\dfrac{1}{{{\log }_{b}}a}+\dfrac{1}{{{\log }_{a}}b}=\sqrt{2020}$. Giá trị của biểu thức $P=\dfrac{1}{{{\log }_{ab}}b}-\dfrac{1}{{{\log }_{ab}}a}$ bằng
A. $\sqrt{2014}$.
B. $\sqrt{2016}$.
C. $\sqrt{2018}$.
D. $\sqrt{2020}$.
A. $\sqrt{2014}$.
B. $\sqrt{2016}$.
C. $\sqrt{2018}$.
D. $\sqrt{2020}$.
Do $a>b>1$ nên ${{\log }_{a}}b>0$, ${{\log }_{b}}a>0$ và ${{\log }_{b}}a>{{\log }_{a}}b$.
Ta có: $\dfrac{1}{{{\log }_{b}}a}+\dfrac{1}{{{\log }_{a}}b}=\sqrt{2020}$
$\Leftrightarrow {{\log }_{b}}a+{{\log }_{a}}b=\sqrt{2020}$
$\Leftrightarrow \log _{b}^{2}a+\log _{a}^{2}b+2=2020$
$\Leftrightarrow \log _{b}^{2}a+\log _{a}^{2}b=2018$ (*)
Khi đó, $P={{\log }_{b}}ab-{{\log }_{a}}ab={{\log }_{b}}a+{{\log }_{b}}b-{{\log }_{a}}a-{{\log }_{a}}b={{\log }_{b}}a-{{\log }_{a}}b$
Suy ra: ${{P}^{2}}={{\left( {{\log }_{b}}a-{{\log }_{a}}b \right)}^{2}}=\log _{b}^{2}a+\log _{a}^{2}b-2=2018-2=2016\Rightarrow P=\sqrt{2016}$
Ta có: $\dfrac{1}{{{\log }_{b}}a}+\dfrac{1}{{{\log }_{a}}b}=\sqrt{2020}$
$\Leftrightarrow {{\log }_{b}}a+{{\log }_{a}}b=\sqrt{2020}$
$\Leftrightarrow \log _{b}^{2}a+\log _{a}^{2}b+2=2020$
$\Leftrightarrow \log _{b}^{2}a+\log _{a}^{2}b=2018$ (*)
Khi đó, $P={{\log }_{b}}ab-{{\log }_{a}}ab={{\log }_{b}}a+{{\log }_{b}}b-{{\log }_{a}}a-{{\log }_{a}}b={{\log }_{b}}a-{{\log }_{a}}b$
Suy ra: ${{P}^{2}}={{\left( {{\log }_{b}}a-{{\log }_{a}}b \right)}^{2}}=\log _{b}^{2}a+\log _{a}^{2}b-2=2018-2=2016\Rightarrow P=\sqrt{2016}$
Đáp án B.